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a, \(A=9x^2-6x+5\)
\(=\left(9x^2-6x+1\right)+4\)
\(=\left(3x-1\right)^2+4\)
ta có:
\(\left(3x-1\right)^2\ge0\forall x\Rightarrow\left(3x-1\right)^2+4\ge4\forall x\)
Vậy Min A = 4
Để A = 4 thì \(3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(b,B=4x^2-5x\)
\(=\left(4x^2-5x+\dfrac{25}{16}\right)-\dfrac{25}{16}\)
\(=\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\)
TA có:
\(\left(2x-\dfrac{5}{4}\right)^2\ge\forall x\Rightarrow\left(2x-\dfrac{5}{4}\right)^2-\dfrac{25}{16}\ge-\dfrac{25}{16}\forall x\)Vậy Min B = \(-\dfrac{25}{16}\)
Để B = \(-\dfrac{25}{16}\) thì \(2x-\dfrac{5}{4}=0\Rightarrow2x=\dfrac{5}{4}\Rightarrow x=\dfrac{5}{8}\)
\(c,C=3x^2-6x\)
\(=3\left(x^2-2x+1\right)-3\)
\(=3\left(x-1\right)^2-3\)
Ta có:
\(3\left(x-1\right)^2\ge0\forall x\Rightarrow3\left(x-1\right)^2-3\ge-3\)
vậy Min C = -3
Để C = -3 thì x-1=0 => x = 1
\(d,D=5x^2-15x\)
\(=5\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{45}{4}\)
\(=5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\)
Ta có:
\(5\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\Rightarrow5\left(x-\dfrac{3}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{45}{4}\)Vậy Min D = \(-\dfrac{45}{4}\)
Để \(D=-\dfrac{45}{4}\) thì \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(e,E=x^2+3x+4\)
\(=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
Vậy Min E = \(\dfrac{7}{4}\) khi \(x+\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(f,F=2x^2-4x+7\)
\(=2\left(x^2-2x+1\right)+5\)
\(=2\left(x-1\right)^2+5\ge5\forall x\)
Vậy Min F = 5 khi x - 1 =0 => x = 1
\(g,2x^2-3x=2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{8}\)
\(=2\left(x-\dfrac{3}{4}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\forall x\)
Vậy Min G = \(\dfrac{-9}{8}\) khi \(x-\dfrac{3}{4}=0\Rightarrow x=\dfrac{3}{4}\)
\(h,H=3x^2-4x=3\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{4}{3}\)
\(=3\left(x-\dfrac{2}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)
Vậy Min H = \(-\dfrac{4}{3}\) khi \(x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
a: \(=-\left(x^2+10x-11\right)\)
\(=-\left(x^2+10x+25-36\right)\)
\(=-\left(x+5\right)^2+36< =36\)
Dấu '=' xảy ra khi x=-5
b: \(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left(x-3\right)^2+4< =4\)
Dấu '=' xảy ra khi x=3
c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
d: \(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9< =9\)
Dấu '=' xảy ra khi x=-1
\(•B=4x-9x^2=-\left(9x^2-4x\right)\\ =-\left(9x^2-3x.2.\dfrac{2}{3}+\dfrac{4}{9}\right)+\dfrac{4}{9}\\ =-\left(3x-\dfrac{2}{3}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\\dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{2}{9}\\ vậy\: MAX_B=\dfrac{4}{9}\: tại\: x=\dfrac{2}{9}\\ •C=5-2x-4x^2=-\left(4x^2+2x-5\right)\\ =-\left(4x^2+2.2x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{21}{4}\\ =-\left(2x+\dfrac{1}{2}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=-\dfrac{1}{4}\\ vậy\: MAX_C=\dfrac{21}{4}\: tại\: x=\dfrac{-1}{4}\\ •D=7+3x-x^2=-\left(x^2-3x-7\right)\\ =-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{37}{4}\\ =-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{3}{2}\\ vậy\: MAX_D=\dfrac{37}{4}\: tại\: x=\dfrac{3}{2}\)\(•E=1+x-x^2=-\left(x^2-x-1\right)\\ =-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\\ dấu\:"="\:xảy\:ra\:khi\:x=\dfrac{1}{2}\\ vậy\:MAX_E=\dfrac{5}{4}\:tại\:x=\dfrac{1}{2}\\ •F=-5x-6x^2\\ -\dfrac{F}{6}=x^2+\dfrac{5}{6}x=x^2+2.x.\dfrac{5}{12}+\dfrac{25}{144}-\dfrac{25}{144}\\ -\dfrac{F}{6}=\left(x+\dfrac{5}{12}\right)^2-\dfrac{25}{144}\\ F=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\\ dấu\: "="\: xảy\: ra\: khi\: x=\dfrac{-5}{12}\\ vậy\: MAX_F=\dfrac{25}{24}\: tại\: x=\dfrac{-5}{12}\)
\(B=4x-9x^2=-9\left(x^2-\dfrac{4}{9}x+\dfrac{4}{81}\right)+\dfrac{4}{9}\)
\(=-9\left(x-\dfrac{2}{9}\right)^2+\dfrac{4}{9}\le\dfrac{4}{9}\forall x\)
vậy Max B = \(\dfrac{4}{9}\) khi \(x-\dfrac{2}{9}=0\Rightarrow x=\dfrac{2}{9}\)
\(C=5-2x-4x^2=-4\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)+\dfrac{21}{4}\)\(=-4\left(x+\dfrac{1}{4}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\)
Vậy Max C = \(\dfrac{21}{4}\) khi \(x+\dfrac{1}{4}=0\Rightarrow x=-\dfrac{1}{4}\)
\(D=7+3x-x^2\)
\(=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{37}{4}\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\forall x\)
Vậy Max D = \(\dfrac{37}{4}\) khi \(x-\dfrac{3}{2}=0\Rightarrow x=\dfrac{3}{2}\)
\(E=1+x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{5}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\forall x\)
Vậy Max E = \(\dfrac{5}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
\(F=-5x-6x^2=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{25}{24}\)\(=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{25}{24}\le\dfrac{25}{24}\forall x\)
Vậy Max F = \(\dfrac{25}{24}\) khi \(x+\dfrac{5}{12}=0\Leftrightarrow x=-\dfrac{5}{12}\)
Bài a,b,c,e,g,i thì đặt điều kiện rồi bình phương 2 vế rồi giải, bài j chuyển vế rồi bình phương
Chỉ trình bày lời giải, tự tìm điều kiện nha :v
d) \(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Rightarrow x-1=1\Leftrightarrow x=2\)
f) \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\sqrt{x-4+2.2\sqrt{x-4}+4}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-4}+2=2\)
\(\Leftrightarrow\sqrt{x-4}=0\)
\(\Rightarrow x-4=0\Leftrightarrow x=4\)
đặt x^2-7x=y=> \(y\ge-\frac{49}{4}\) (*)
\(A=y\left(y+12\right)=y^2+12y=\left(y+6\right)^2-36\ge-36\)
đẳng thức khi y=-6 thủa mãn đk (*)
Vậy: GTNN của A=-36 khí y=-6 =>\(\left[\begin{matrix}x=1\\x=6\end{matrix}\right.\)
\(A=x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\)
Vậy \(A_{min}=1\Leftrightarrow x=-1\)
\(B=x^2+4x=6=x^2+4x+4+2=\left(x+2\right)^2+2\ge2>0\)
Vậy \(B_{min}=2\Leftrightarrow x=-2\)
a) \(A=x^2-2x-6\)
\(A=\left(x^2-2x+1\right)-7\)
\(A=\left(x-1\right)^2-7\)
Mà \(\left(x-1\right)^2\) luôn \(\ge\)\(0\) => GTNN của biểu thức là -7 với \(\left(x-1\right)^2=0\) tức x=1
a: \(=x^2-2x+1-7=\left(x-1\right)^2-7>=-7\)
Dấu '=' xảy ra khi x=1
b: \(=4x^2-4x+1+6=\left(2x-1\right)^2+6>=6\)
Dấu '=' xảy ra khi x=1/2
c: \(=9x^2-6x+1-1=\left(3x-1\right)^2-1>=-1\)
Dấu '=' xảy ra khi x=1/3
d: \(=x^2+12x+36-36=\left(x+6\right)^2-36>=-36\)
Dấu '=' xảy ra khi x=-6
e: \(=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}>=-\dfrac{9}{4}\)
Dấu '=' xảy ra khi x=3/2